MHB How to prove this logarithmic inequality?

AI Thread Summary
The discussion focuses on proving the inequality that states (log_{24}(48))^2 + (log_{12}(54))^2 > 4. The original poster struggles with the problem, attempting to convert logarithms to base-10 and express the numbers in terms of 2 and 3, but finds it confusing. A participant suggests a simpler approach by breaking down the logarithms: log_{24}(48) can be expressed as 1 + log_{24}(2), and log_{12}(54) as 1 + log_{12}(4.5). This leads to the conclusion that the inequality holds true by demonstrating that the squared terms exceed 4. The discussion emphasizes the importance of simplifying logarithmic expressions to prove inequalities effectively.
anemone
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Hi all, I've been having a hard time trying to solve the following inequality:

Prove that $\displaystyle \left(\log_{24}(48) \right)^2+\displaystyle \left(\log_{12}(54) \right)^2 >4$

I've tried to change the bases to base-10 log and relating all the figures (12, 24, 48, and 54) in terms of 2 and 3 but only to make the problem to be more confounded.

Could I get some hints on how to tackle this problem?

Any help would be deeply appreciated.

Thanks!

P.S. This question was originally asked here (http://www.mymathforum.com/viewtopic.php?f=13&t=27644&p=110515&hilit=noki#p110515) at MMF.
 
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anemone said:
Hi all, I've been having a hard time trying to solve the following inequality:

Prove that $\displaystyle \left(\log_{24}(48) \right)^2+\displaystyle \left(\log_{12}(54) \right)^2 >4$

I've tried to change the bases to base-10 log and relating all the figures (12, 24, 48, and 54) in terms of 2 and 3 but only to make the problem to be more confounded.

Could I get some hints on how to tackle this problem?

Any help would be deeply appreciated.

Thanks!

P.S. This question was originally asked here (http://www.mymathforum.com/viewtopic.php?f=13&t=27644&p=110515&hilit=noki#p110515) at MMF.

\(\log_{24}(48)=1+\log_{24}(2)\)

But \(2^5 \gt 24\) so \(\log_{24}(2) \gt 1/5\)

Also: \(\log_{12}(54)=1+\log_{12}(4.5)\), and \(4.5^5>12^3\) so \(\log_{12}(4.5)>3/5\)

Hence:
\[ (\log_{24}(48))^2 + (\log_{12}(54))^2 \gt 1.2^2+1.6^2 =4 \]

CB
 
Hi CB, a big thank for your help in making it so straightforward and simple for me!

Thanks.
 
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